Physics 140. Sound. Chapter 12

Transcription

1 Physics 140 Sound Chapter 12

2 Sound waves Sound is composed of longitudinal pressure waves. wave propagabon Compression Compression Compression è when parbcles come together RarefacBon RarefacBon RarefacBon è when parbcles are farthest apart 2

3 Sound waves Described by the gauge pressure difference between the pressure at a given point and average pressure 3

4 The speed of sound waves Let s remember Chapter 11 and the speed of string waves v T µ The speed of sound waves v B ρ B is the bulk modulus of the medium and ρ its density. The speed of sound waves in thin rods v Y ρ Y is the Young s modulus of the medium and ρ its density. All other equabons from Chapter 11 sbll apply. v f λ k 2π λ 4

5 Conceptual quesbon Speed of sound Q1 Given the informabon in the table below, which substance would have the highest speed of sound? Substance Bulk Modulus Density (GPa) (kg/m 3 ) Aluminum Speed (m/s) Brass Copper Mercury A. aluminum B. brass C. copper D. mercury v B ρ 5

6 Conceptual quesbon Speed of sound Q1 Given the informabon in the table below, which substance would have the highest speed of sound? Substance Bulk Modulus Density Speed (GPa) (kg/m 3 ) (m/s) Aluminum x 10 3 Brass x 10 3 Copper x 10 3 Mercury x 10 3 A. aluminum B. brass C. copper D. mercury v B ρ 6

7 Exercise: Speed of sound in gold What s the speed of sound in a gold ingot? Density of gold is ρ 19,360 kg/m 3 and its bulk modulus is 220 GPa. v B ρ What s given: ρ 19,360 kg/m 3 B 220 GPa 220x10 9 Pa v B ρ v Pa kg/m m/s 7

8 CharacterisBcs of sound Sound can travel through any kind of ma`er, but not through a vacuum. The speed of sound is different in different materials; in general, it is slowest in gases, faster in liquids, and fastest in solids. The speed depends somewhat on temperature, especially for gases. v v 0 T T 0 V 0 is the absolute speed at absolute temperature T 0. e.g. v 0 (air) 343 m/s at T C. 8

9 The speed of sound waves Speed of sound in ideal gases: v v 0 T T 0 SI units of temperature: [K] To convert temperature from 0 C to Kelvins (SI unit of temperature) you just need to add 273 K. T(in K) T C (in C)+ 273 For the speed of sound in air (and only in air) we can use an approximate formula: v ( T C )m/s where T c is the air temperature in 0 C. 9

10 Exercise: A lightning flash A lightning flash is seen in the sky and 8.2 seconds later the boom of thunder is heard. The temperature of the air is C. (a) What is the speed of sound in air at that temperature? What is given: t 8.2 s T C 12 0 C Since we are looking for the speed of sound in air, we can use the approximate formula: v ( T C )m/s v (b) How far away is the lightning strike? ( C) m/s 338 m/s d vt d ( 338 m/s)( 8.2s) 2800m 2.8km (c) The speed of light is 3.00x10 8 m/s. How long does it take the light signal to reach the observer? t d v 2800 m 6 t s km/s 10

11 The amplitude and intensity The intensity of a wave is the energy transported per unit Bme across a unit area. It can be given as a funcbon of a pressure amplitude (P 0 ) I P 2 0 2ρv Or by looking at the amplitude of the displacement of parbcles P 0 ωvρs max ω is the angular frequency of sound v is the speed of sound ρ is the density of the medium s max is the maximum displacement Units of intensity: W/m 2 11

12 Loudness and decibels Loudness of a sound is measured by the logarithm of the intensity. Sound level (intensity level) is measured in decibels (db) and is defined: β ( 10 db) log I I 0 I 0 is taken to be the threshold of hearing: I W/m 2 Sound level is not the same as the intensity of the sound wave! 12

13 Exercise: Decibels from a speaker The sound level 25 m from a loudspeaker is 71 db. a) What is the rate at which sound energy is being produced by the loudspeaker, assuming it to be an isotropic source? What is given: r 25 m β 71 db β ( 10 db) log I I 0 I W/m 2 We can solve for the intensity of the sound wave: β ( 10dB) log I I 0 log I β I 0 10dB I 10 I 0 β 10dB I W/m dB 10 db W/m 2 13

14 Exercise: Decibels from a speaker The sound level 25 m from a loudspeaker is 71 db. a) What is the rate at which sound energy is being produced by the loudspeaker, assuming it to be an isotropic source? What is given: r 25 m β 71 db And we calculated: I 1.3x10-5 W/m 2 We need to find the rate at which sound energy is being produced that s power, but we know intensity of sound and distance. OK, let s go back to Chapter 11 and use the equabon for intensity of the isotropic source: I P 2 P I4πr 4πr 2 P ( W/m 2 )4π 2 ( 25 m) 0.10 W 14

15 Exercise: Decibels from a speaker The sound level 25 m from a loudspeaker is 71 db. b) What is the pressure amplitude of a sound wave? Density of air is 1.2 kg/m 3 and the speed of sound is about 343 m/s. What is given: r 25 m β 71 db ρ 1.2 kg/m 3 v 340 m/s And we calculated: I 1.3x10-5 W/m 2 I P 2 0 2ρv P0 I 2ρv P W/m kg/m 343 m/s 0.1Pa 15

16 Standing waves of sound All of these musical instruments are effecbvely sealed off on one end and open on the other. The air molecules right next to the closed end are essenbally not moving è displacement nodes The molecules at the open end are exposed to the atmosphere. This gives them the greatest freedom of movement è displacement anbnodes 16

17 Standing waves of sound - Tube closed at one end - L First harmonic - fundamental L λ 1 4 λ 1 4L Second harmonic L 3λ 1 4 λ L Third harmonic L 5λ 1 4 λ L 17

18 Standing waves of sound - Tube closed at one end - The general result for standing waves in a tube open at one end and closed at the other is λ n 4L n where n 1, 3, 5, (odd values only!!) f n v λ n nv 4L nf 1 f 1 is the fundamental frequency. Or λ n' 4L 2n' 1 where n 1, 2, 3, Frequencies have only odd mulbples of the fundamental one è Even harmonics are not present 18

19 Exercise: A bugle What are the first 3 harmonic frequencies emi`ed by a bugle that has a tube 50 cm long arer it was ler out in the cold, cold night (assume the temperature is -12 o C)? What is given: L 50 cm 0.5 m T C C f n (2n 1)v 4L (2n 1) f 1 Since we are looking for the speed of sound in air, we can use the approximate formula: v ( T C )m/s ( ( 0 v C ) m/s m/s n 1 - fundamental frequency: f 1 (2n 1)v 4L v 4L f m/s m Hz n 2 f 2 (2 2 1) f 1 3 f 1 f Hz f ( f f Hz ) f 5 n

20 Pressure variabons - Tube closed at one end - Displacement node at the closed end corresponds to the pressure anbnode è molecules at that locabon are the ones gesng maximally compressed Displacement anbnode at the open end corresponds to the pressure node è molecules at that locabon are the ones gesng maximally stretched out 20

21 Standing waves of sound - Tube open at both ends - L First harmonic - fundamental L λ 1 2 λ 1 2L Second harmonic L 2λ 1 2 λ 1 L Third harmonic L 3λ 1 2 λ L 21

22 Standing waves of sound - Tube opened at both ends - The general result for standing waves in a tube open at one end and closed at the other is λ n 2L n f n v λ n nv 2L nf 1 f 1 is the fundamental frequency. Unlike the pipe open at only one end, all harmonics are present. Since both ends are open, air molecules at each end have maximum possible displacements è displacement anbnodes 22

23 Pressure variabons - Tube opened at both ends - Displacement anbnodes at the open ends correspond to the pressure nodes è molecules at that locabon are the ones gesng maximally stretched out 23

24 Exercise: An organ pipe An organ pipe that is open at both ends has a fundamental frequency of 382 Hz at 0 0 C. a) What is the fundamental frequency for this pipe at 20 0 C? What is given: f 1,0 382 Hz T C T C f n v λ n nv 2L nf 1 f 1 v 2L The speed of sound in air at: T C is v m/s T C is v m/s Note that in case you need to find the speed of sound at any other temperature not provided in the table, you would need to use the eqn. for the speed of sound (Slides 8 and 9). The fundamental frequency at 0 0 C: The fundamental frequency at 20 0 C: f f 1,0 f 1,20 v0 2L v20 2L 343 m/s 382 Hz 331m/s 1, Hz We can find the rabo: f 1,20 f 1,0 v20 2L v0 2L v v

25 Exercise: An organ pipe An organ pipe that is open at both ends has a fundamental frequency of 382 Hz at 0 0 C. b) How long is the pipe? What is given: f 1,0 382 Hz T C T C f 1,0 v0 2L 2L v f 0 1,0 331Hz L m/s 0.43 m You could also choose the other equabon, for f 1,20 and v 20 and sbll get the same answer. 25

26 Timbre Standing wave is a superposibon of many standing wave pa`erns with different frequencies Fundamental frequency (1 st harmonic) Overtones (higher harmonics) The same note sounds differently on different instruments è fundamental frequency might be the same but the overtones appear in different intensibes Timbre (tone quality) 27

27 Beats Let s consider two different waves, each with the same wave speed but with slightly different frequencies (and, hence, wavelengths). The superposibon of the waves will produce a pulsabon called beats. 28

28 Beats Beat frequency is Δf f 1 f 2 If the beat frequency exceeds about 15 Hz, the ear will perceive two different tones instead of beats. Example: A violin is tuned by adjusbng the tension in the strings. Brian s A string is tuned to a slightly lower frequency than Jennifer s, which is correctly tuned to 440 Hz. What is the frequency of Brian s string if beats of 2 Hz are heard when they bow the strings together? Δf f 1 f 2 f 2 f 1 Δf 438Hz What would happen if the temperature where Brian seats drops from 25 0 C to 20 0 C? The frequency of Brian s violin would change: v v 0 ( ) f Hz ( ) T T 0 f f 0 T T 0 29

29 The Doppler effect The Doppler effect occurs when a source of sound is moving with respect to an observer. When source is moving toward an observer the sound has a higher frequency and shorter wavelength When a source is moving away from an observer the sound has a lower frequency and longer wavelength 30

30 The Doppler effect v sound f λ 1 T λ λ v sound T Crest The observer will hear a sound wave with a different wavelength: λ 0 λ d λ v source T λ 0 λ v source λ v sound In the same Bme T, the source will move distance d toward the observer: d v source T " λ 0 λ 1 v % source $ ' # & v sound 31

31 The Doppler effect In general, when both the observer and the source are moving:! f 0 v ± v sound observer # " v sound ± v source $ & f % When the observer is moving towards the source use v sound + v observer When the observer is moving away from the source use v sound - v observer When the source is moving towards the observer use v sound v source When the source is moving away from the observer use v sound + v source Note, that if the observer or the source are stabonary, just set their speeds to zero. 32

32 Exercise: Rocked sleds A maniac on a rocket sled is moving along a train track. She hears a train whistle at a frequency of 1200 Hz. The maniac, being a big fan of trains, knows that a train travels at an average speed of 30 m/s and the whistle is normally at a frequency of 720 Hz. What is the speed of the maniac if she starts sledding towards the train? Assume that the speed of sound is 340 m/s Hz! f 0 v ± v sound observer # " v sound ± v source $ & f % 720 Hz 340 m/s 30 m/s Since the observer (maniac) is moving towards the source (train): v sound + v observer Since the source (train) is moving towards the observer (maniac): v sound v source " f 0 v sound + v observer % $ ' f # v sound v source & 33

33 Exercise: Rocked sleds A maniac on a rocket sled is moving along a train track. She hears a train whistle at a frequency of 1200 Hz. The maniac, being a big fan of trains, knows that a train travels at an average speed of 30 m/s and the whistle is normally at a frequency of 720 Hz. What is the speed of the maniac if she starts sledding towards the train? " f 0 v sound + v $ observer # v sound v source v sound % ' f & f f 0 + vobserver v sound + v observer f 0 v sound v source f ( v v ) sound source 1200 Hz 340 m/s+ v observer 720 Hz ( 340 m/s 30 m/s) m/s v observer m/s 34

34 EcholocaBon Sound waves can be sent out from a transmi`er of some sort; they will reflect off any objects they encounter and can be received back at their source. The Bme interval between emission and recepbon can be used to build up a picture of the scene. If the sound takes Bme t to go from the source (bat, dolphin, ) to the object and back then it travels the distance: d v sound t 2 And all equabons for calculabng the speed of sound sbll apply! 35

35 Exercise: A sonar A boat is using sonar to detect the bo`om of a freshwater lake. If the echo from a sonar signal is heard s arer it is emi`ed, how deep is the lake? Assume the lake s temperature is uniform and at 25 0 C. What is given: t 0.54 s T 25 0 C d v sound t 2 From table 12.1 in our textbook the speed of sound in freshwater is 1493 m/s. d 1493 m/s 0.27 s m 36

36 How do I solve for the logarithm: β ( 10dB) log I I 0 log I β I 0 10dB I 10 I 0 β 10dB General rule: log B A C A B C 37

37 Summary The speed of sound waves v B ρ B is the bulk modulus of the medium and ρ its density. The speed of sound waves in thin rods v Y ρ Y is the Young modulus of the medium and ρ its density. v f λ k 2π λ The speed depends v v0 temperature - gases 0 T T V 0 is the absolute speed at absolute temperature T 0. 38

38 Summary Intensity if a sound wave as a funcbon of a pressure amplitude I P 0 2ρv 2 P 0 ωvρs max Sound level (intensity level) β ( 10dB) log I I T(in K) T C (in C)+ 273 Beat frequency is 0 Δf I W/m 2 f 1 f 2 The Doppler effect! f 0 v sound ± v # observer " v sound ± v source $ & f % 39